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co-ordinate -slope-intercept problem

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by kishokbabu » Tue Dec 13, 2011 11:22 am
A straight line passes through the points (a,b) & (c,d). Is the slope of the line less than 0?

1) (a-c)(b-d)<0
2) Product of the intercepts of the line on the X-axis and on the Y-axis is greater than 0.

Pls explain how to solve this problem?
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Source: — Data Sufficiency |
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by neelgandham » Tue Dec 13, 2011 2:04 pm
A straight line passes through the points (a,b) & (c,d). Is the slope of the line less than 0?
1)(a-c)(b-d)<0
Slope of a line = (y2-y1)/(x2-x1) where (x1,y1) and (x2,y2) are points on the line
Slope of the line above = (b-d)/(a-c)
Given (a-c)(b-d)<0 implies, either (a-c)>0 and (b-d)<0 OR (a-c)<0 and (b-d)>0. In both the cases
Slope of the line above = (b-d)/(a-c) < 0
Sufficient!
2)Product of the intercepts of the line on the X-axis and on the Y-axis is greater than 0
.

Equation of the line => y = mx + c, where m is the slope of the line and c is the y intercept
x intercept of the line(y=0) => x = -c/m
Product of x intercept and y intercept = -(c^2)/m > 0(From statement 2)
-(Square of a number)/m >0 implies
-ve * +ve/m = +ve
so m = -ve < 0
Sufficient to answer the question Is the slope of the line less than 0?

IMO Answer D
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by Ian Stewart » Tue Dec 13, 2011 2:16 pm
kishokbabu wrote:A straight line passes through the points (a,b) & (c,d). Is the slope of the line less than 0?

1) (a-c)(b-d)<0
2) Product of the intercepts of the line on the X-axis and on the Y-axis is greater than 0.

Pls explain how to solve this problem?
The question asks if the slope of the line is negative. A line has a negative slope if it is falling as it moves to the right.

From Statement 1, we learn that (a-c) and (b-d) have opposite signs. Now the slope of the line is (b-d)/(a - c) (using the standard slope formula); if (a-c) and (b-d) have opposite signs, then this fraction must be negative, so Statement 1 is sufficient.

From Statement 2, we learn that the x-intercept and y-intercept of the line have the same sign: both are positive or both are negative. If both are positive, then the line crosses the y-axis somewhere above the origin (0,0). For it to then have a positive x-intercept, it must fall as it moves to the right, and thus has a negative slope. Similarly, if the x-intercept is negative, the line crosses the x-axis to the left of the origin. For the line to then have a negative y-intercept, the line must again fall as it moves to the right, so thus has a negative slope. All of this is easier to see if you sketch quick diagrams of course. So Statement 2 is sufficient.

The answer is D.
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